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 2 years ago
[SOLVED] Here's a cute predicate logic problem.
Prove or disprove: \[\forall x \,\exists y \,\forall z\, \exists w \;\;\text{such that} \;\;x+y+z+w=xyzw\]Assume that \(x, y, z, w \in \mathbb{C}\).
 2 years ago
[SOLVED] Here's a cute predicate logic problem. Prove or disprove: \[\forall x \,\exists y \,\forall z\, \exists w \;\;\text{such that} \;\;x+y+z+w=xyzw\]Assume that \(x, y, z, w \in \mathbb{C}\).

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experimentX
 2 years ago
Best ResponseYou've already chosen the best response.0is xyzw a combination of digit or multiplication of digit??

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.0lol .. didn't realize they were complex

KingGeorge
 2 years ago
Best ResponseYou've already chosen the best response.3It doesn't really matter. I just put that specification on there so that know one tried to change the problem so that addition and multiplication had different meanings.

KingGeorge
 2 years ago
Best ResponseYou've already chosen the best response.3Just prove/disprove it for \(\mathbb{R}\) and it will become clear why it doesn't matter if they're complex numbers.

KingGeorge
 2 years ago
Best ResponseYou've already chosen the best response.3Hint below in white LaTeX\[\color{white}{\text{There is a certain value for y that you should choose to prove it.}}\]

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.0\[ \color{white}{\text{LOL i didn't know that!!.}} \]

KingGeorge
 2 years ago
Best ResponseYou've already chosen the best response.3Another, probably more helpful hint, below in white LaTeX\[\color{white}{\text{What value of y can you choose so that xyzw is constant?}}\]

nbouscal
 2 years ago
Best ResponseYou've already chosen the best response.2Choose y=0, then choose w=(x+z).

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.0But ∃y doesn't mean that y is rather a variable such that we have to fit constraint ??

KingGeorge
 2 years ago
Best ResponseYou've already chosen the best response.3\(\exists\, y\) means that we can choose a specific value for y. It's a variable, but one that we have complete control over.

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.0then it must be false ... i guess so!!

KingGeorge
 2 years ago
Best ResponseYou've already chosen the best response.3The "there exists" only means that at least one possible value of y where this is true. One value that we know works is \(y=0\). This means that there does exists at least one value for \(y\) that makes the statement true, so we have proven that the original statement is true.

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.0does't y=0 make statement untrue?? xyzw=0, for y=0

nbouscal
 2 years ago
Best ResponseYou've already chosen the best response.2Right, both the sum and the multiplication end up equaling zero because you set w=(x+z).

KingGeorge
 2 years ago
Best ResponseYou've already chosen the best response.3If \(y\) is 0, and \(x=(x+z)\) we have \[x+0+z+((x+z))=x+zxz=0\]and \[x\cdot0\cdot z\cdot w=0\]So that means that \[x+y+z+w=xyzw\]

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.0Oh ... stupid me ... i never considered the negative values!!

KingGeorge
 2 years ago
Best ResponseYou've already chosen the best response.3Hope everyone enjoyed this little (unintuitive) problem. Good night.

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.0thanks!! i'll try to think broader way next time!!

KingGeorge
 2 years ago
Best ResponseYou've already chosen the best response.3I'll try to get a problem just for you next time then. Any preference about the subject?
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